Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $x = \dfrac{9y - 54}{8y^2 + 8y} \times \dfrac{y^2 + 7y + 6}{-3y + 18} $
First factor the quadratic. $x = \dfrac{9y - 54}{8y^2 + 8y} \times \dfrac{(y + 1)(y + 6)}{-3y + 18} $ Then factor out any other terms. $x = \dfrac{9(y - 6)}{8y(y + 1)} \times \dfrac{(y + 1)(y + 6)}{-3(y - 6)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ 9(y - 6) \times (y + 1)(y + 6) } { 8y(y + 1) \times -3(y - 6) } $ $x = \dfrac{ 9(y - 6)(y + 1)(y + 6)}{ -24y(y + 1)(y - 6)} $ Notice that $(y - 6)$ and $(y + 1)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 9\cancel{(y - 6)}(y + 1)(y + 6)}{ -24y\cancel{(y + 1)}(y - 6)} $ We are dividing by $y + 1$ , so $y + 1 \neq 0$ Therefore, $y \neq -1$ $x = \dfrac{ 9\cancel{(y - 6)}\cancel{(y + 1)}(y + 6)}{ -24y\cancel{(y + 1)}\cancel{(y - 6)}} $ We are dividing by $y - 6$ , so $y - 6 \neq 0$ Therefore, $y \neq 6$ $x = \dfrac{9(y + 6)}{-24y} $ $x = \dfrac{-3(y + 6)}{8y} ; \space y \neq -1 ; \space y \neq 6 $